(related to Lemma: Subgroups of Cyclic Groups)
Let \((G,\ast)\) be a cyclic group generated by \(g\), i.e. \(G=\langle g \rangle\) and let \(H\subseteq G\) be its subgroup. We can easily recognize trivial subgroups \(H=\{e\}=\langle g^0 \rangle\) and \(H=G=\langle g^1 \rangle\) as cyclic. Thus, assume \(H\) being another subgroup of \(G\). In such a case, \(H\) must contain at least one element \(x=g^k\) with a natural number \(k > 1\). Consider the set of all such natural numbers, more precisely
\[K:=\{k~|~x=g^k, x\in H,~ k > 1\}.\]
Because \(H\) is non-empty by assumption, also \(K\) is non-empty. By virtue of the well-ordering principle, the set \(K\) contains a smallest element, call it \(b\). Therefore, also \(H\) contains the element \(g^b\). Because \(H\) is closed under the operation \(\ast\), it must contain all powers of \((g^b)^q=g^{b\cdot q}\), \(q\in\mathbb Z\) but no other powers of \(g\). For if it contained a power of \(g\), e.g. \(g^a\), whose exponent \(a\) was not a multiple of \(b\), then we could use a unique division with a quotient and remainder to write \[a=q\cdot b+r\] with \(0 < r < b\). In this case, however, it would follow that \(g^r\in H\), and so that \(r\in K\), in contradiction to the minimality of \(b\) in \(K\). Therefore, \(H\) is cyclic and generated by \(g^b\), i.e. \(H=\langle g^b \rangle\).